all I could think about was to use the fact that if a sequence is convergent then all of its subsequences that converge must covnerge to the same limit of the sequence itself; maybe I could assume that these subsequences converge to different limits and get a contradiction?
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AnonymousApr 30 '12 at 17:50

Yes, you could approach it by contradiction, but there's an easier way. Let $L_{\text{odd}}=\lim_n a_{2n+1}$ and $L_{\text{even}}=\lim_n a_{2n}$, and use the convergence of $\langle a_{3n}\rangle$ to show that $L_{\text{odd}}=L_{\text{even}}$.
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Brian M. ScottApr 30 '12 at 17:53

2 Answers
2

Hint: first show that the three subsequences have the same limit. (The subsequence $(a_{3n})$ has a further subsequence that is a subsequence of $(a_{2n})$, for instance.)

Then note that given $n>1$, $a_n$ is a term of one of the subsequences $(a_{2n})$, $(a_{2n+1})$. (So, given $\epsilon>0$, choose $N$ so that for any $n\ge N$, each of $a_{2n}$ and $a_{2n+1}$ is within $\epsilon$ of the common limit. Then... .)

"The subsequence $(a_{3n})$ has a further subsequence that is a subsequence of $(a_{2n})$" - that's clear; however, why can I deduce that therefore $(a_{3n})$ and $(a_{2n})$ have the same limit?
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AnonymousApr 30 '12 at 18:06

@Anonymous Suppose $(a_{2n})$ converges to $L$. Then each of its subsequences converges to $L$. So some subsequence of $(a_{3n})$ converges to $L$. So $(a_{3n})$ converges to $L$ (since $(a_{3n})$ is convergent).
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David MitraApr 30 '12 at 18:09

OK, awesome; therefore all the three subsequences have the same limit. let $n>1$, I understand that $a_n$ is a term of the subsequences $(a_{2n})$, $(a_{2n+1})$. how do I continue from here? I know that if n is even I can write: $|a_{2n}-L|<\epsilon$ and if n is odd then $|a_{2n+1}-L|<\epsilon$; how can I deduce now that $|a_n-L|<\epsilon$?
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AnonymousApr 30 '12 at 18:17

1

@Anonymous Finishing up the parenthetical comment at the end of my answer: For $n\ge N$, both $a_{2n}$ and $a_{2n+1}$ are within $\epsilon$ of $L$. But then every term $a_{2N}$, $a_{2N+1}$, $a_{2N+2}$, $a_{2N+3}$, $\ldots\,$ is within $\epsilon$ of $L$. Note those latter terms are a tail of $(a_n)$; so, for the sequence $(a_n)$, if $n\ge 2N$ ... .
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David MitraApr 30 '12 at 18:32

finally got it; as always clear as crystal; thanks again for your great help.
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AnonymousApr 30 '12 at 18:49